Question

Consider the so called D-T reaction (deuterium-tritium fusion) `._1H^2+._1H^3to._2He^4+n`
Calculate the energy released in MeV in this reaction from the data
`m(._1H^2)=2.014102u, m(._1H^3)=3.016049u`
(b) Consider the radius of both deuterium and tritium to be approximately 2.0fm. what is the kinetic energy needed to overcome the Coulomb repulsion between the two nuclei? To what temperature must the gases the be heated to initiate the reaction?

Answer 1

(a) Take the D-T nuclear reaction : `""_(1)^(2)H+""_(1)^(3)H rarr ""_(2)^(4)He+n`
It is given that:
Mass of `""_(1)^(2)H,m_(1)=2.014102 u`
Mass of `""_(1)^(3)H,m_(2)= 3.016049 u `
Mass of `""_(2)^(4)H,m_(3)=4.002603u`
Mass of `""_(0)^(1)n,m_(4)=1.008665 u`
Q-value of the given-T reaction is :
`Q=[m_(1)+m_(2)-m_(3)-m_(4)]c^(2)`
`=[2.014102+3.016049-4.002603-1.008665]c^(2)`
`=[0.018883 c^(2)]u`
But 1 u=931.5 MeV/`c^(2)`
`therefore Q=0.018883xx931.5=17.59 MeV`
(b) Radius of deuterium and tritium , `r~~2.0 " " fm=2xx10^(-15)m`
Distance between the two nuclei at the moment when they touch each other,d `=r+r=4xx10^(-15)m`
Charge on the deuterium nucleus=e
Charge on the tritium nucleus=e
Hence, the repulsive potential energy between the two nuclei is given as :
`V=(e^(2))/(4pt in_(0)(d))`
Where, `in_(0)`= Permittivity of free space
`(1)/(4 pi in_(0))=9xx10^(9)Nm^(2)C^(-2)`
`therefore V=(9xx10^(9)xx(1.6xx10^(-19))^(2))/(4xx10^(-15))=5.76xx10^(-14)J`
`(5.76xx10^(-14))/(1.6xx10^(-19))=3.6xx10^(5) eV=360keV`
Hence, `5.76xx10^(-14)J` or 360keV of kinetic energy (KE) is needed to overcome the Coloumb repulsion between the two nuclei.
However, it is given that:
`KE=2xx(3)/(2)kT`
Where, k=Boltzmann constant`=1.38xx10^(-23) m^(2)kg s^(-2)K^(-1)`
T= Temperature rquired for triggering the reaction
`thereforeT=(KE)/(3K)`
`=(5.76xx10^(-14))/(3xx1.38xx10^(-23))=1.39xx10^(9) K`
Hence, the gas must be heated to a temperature of `1.39xx10^(9)`K t initiate the reaction.